哪位大哥帮忙解解这道高一数学题
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2013-08-25 · 知道合伙人教育行家
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现在为上海海事大学学生，在学习上有一定的经验，擅长数学。
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（1）原式
=[2cos(30-20)-sin20]/cos20
=(2cos30cos20+2sin30sin20-sin20]/cos20
=(2*√3/2*cos20+2*1/2sin20-sin20)/cos20
=(√3cos20+sin20-sin20)/cos20
=√3cos20/cos20
=√3
（2）原式
=(cosπ/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11*2sinπ/11)/2sinπ/11
=(sin2π/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11)/2sinπ/11
=(1/2*sin4π/11*cos3π/11*cos4π/11*cos5π/11)/2sinπ/11
=(1/4*sin8π/11*cos3π/11*cos5π/11)/2sinπ/11
=(1/4*sin3π/11*cos3π/11*cos5π/11)/2sinπ/11
=(1/8*sin6π/11*cos5π/11)/2sinπ/11
=(1/8*sin5π/11*cos5π/11)/2sinπ/11
=(1/16*sin10π/11)/2sinπ/11
=1/32
=[2cos(30-20)-sin20]/cos20
=(2cos30cos20+2sin30sin20-sin20]/cos20
=(2*√3/2*cos20+2*1/2sin20-sin20)/cos20
=(√3cos20+sin20-sin20)/cos20
=√3cos20/cos20
=√3
（2）原式
=(cosπ/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11*2sinπ/11)/2sinπ/11
=(sin2π/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11)/2sinπ/11
=(1/2*sin4π/11*cos3π/11*cos4π/11*cos5π/11)/2sinπ/11
=(1/4*sin8π/11*cos3π/11*cos5π/11)/2sinπ/11
=(1/4*sin3π/11*cos3π/11*cos5π/11)/2sinπ/11
=(1/8*sin6π/11*cos5π/11)/2sinπ/11
=(1/8*sin5π/11*cos5π/11)/2sinπ/11
=(1/16*sin10π/11)/2sinπ/11
=1/32
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